# Distinguished Opens (Tuesday, September 22)

## Computing the Regular Functions

Given an affine variety $X$ and $U\subseteq X$ open, a *regular function* $\phi: U\to k$ is one locally (wrt the Zariski topology) a fraction.
We write the set of regular functions as $\OO_X$.

:::{.example}
$X = V(x_1 x_4 - x_2 x_3)$ on $U = V(x_2, x_4)^c$, the following function is regular:
\[  
\phi: U &\to k \\
x &\mapsto 
\begin{cases}
{x_1\over x_2} & x_2 \neq 0 \\ \\
{x_3 \over x_4} & x_4 \neq 0
\end{cases}
.\]
Note that this is not globally a fraction.
:::

## Distinguished Opens

:::{.definition title="Distinguished Open Sets"}
A *distinguished open set* $D(f) \subseteq X$ for some $f\in A(X)$ is $V(f)^c \da \ts{x\in X \st f(x) \neq 0}$.
:::

These are useful because the $D(f)$ form a base for the Zariski topology.

:::{.proposition title="?"}
For $X$ an affine variety, $f\in A(X)$, we have
\[ 
\OO_X(D(f)) = \ts{ {g\over f^n} \st g\in A(X), n\in \NN}
.\]
:::

:::{.proof}
The first reduction we made was that $\phi \in \OO_X(D(f))$ is expressible as $g_a /  f_a$ on distinguished opens $D(f_a)$ covering $D(f)$.
We also noted that 
\[
{g_a \over f_a} = {g_b \over f_b} \text{ on } D(f_a) \intersect D(f_b) \implies f_b g_a = f_a g_b \text{ in } A(X)
.\]
\

The second step was writing $D(f) = \union D(f_a)$, and so $V(f) = \intersect_a V(f_a)$ implies that $f\in I\qty{ V\qty{ \ts{f_a \st a\in U} }}$.
By the Nullstellensatz, $f\in \sqrt{\gens{f_a \st a\in U}}$, so $f^N = \sum k_a f_a$ for some $N$.
So construct $g = \sum k_a g_a$, then compute
\[  
gf_b = \sum_a k_a g_a f_b = \sum_a k_a g_b f_a = g_b \sum k_a f_a = g_b f^N
.\]
Thus $g/f^N = g_b / f_b$ for all $b$, and we can thus conclude
\[  
\phi \da \ts{{g_b \over f_b} \text{ on } D(f_b)} = g/f^N
.\]
:::

:::{.corollary title="?"}
For $X$ an affine variety, $\OO_X(X) = A(X)$.
:::

:::{.warnings title="Things go wrong when $k\neq \bar k$"}
For $k$ not algebraically closed, the proposition and corollary are both false.
Take $X = \AA^1/\RR$, then ${1\over x^2+1} \in \RR(x)$, but $\OO_X(X) \neq A(X) = \RR[x]$.
:::

## Structure Sheaf of Distinguished Opens

:::{.definition title="Localization"}
Let $R$ be a ring and $S$ a set closed under multiplication, then the localization at $S$ is defined by
\[  
R_S \da \ts{r/s \st r\in R, s\in S} / \sim
.\]
where $r_1/s_1 \sim r_2/s_2 \iff s_3(s_2 r_1 - s_1 r_2) = 0$ for some $s_3 \in S$.
:::

:::{.example}
Let $f\in R$ and take $S = \ts{f^n \st n\geq 1}$, then $R_f \da R_S$.
:::


:::{.corollary title="?"}
$\OO_X(D(f)) = A(X)_f$ is the localization of the coordinate ring.
:::

These requires some proof, since the LHS literally consists of functions on the topological space $D(f)$ while the RHS consists of formal symbols.

:::{.proof}
Consider the map
\[  
A(X)_f &\to \OO_X(D(f)) \\
``g/f^n" &\mapsto g/f^n: D(f) \to k
.\]

By definition, there exists a $k\geq 0$ such that 
\[  
f^k(f^m g - f^n g') = 0 
\implies
f^k(f^m g - f^n g') = 0 \text{ as a function on } D(f)
.\]
Since $f^k \neq 0$ on $D(f)$, we have $f^m g = f^n g'$ as a function on $D(f)$, so $g/f^n = g'/g^m$ as functions on $D(f)$.
\

**Surjectivity**: By the proposition, we have surjectivity, i.e. any element of $|OO_x(D(f))$ can be represented by some $g/f^n$.
\

**Injectivity**:
Suppose $g/f^n$ defines the zero function on $D(f)$, then $g = 0$ on $D(f)$ implies that $fg=0$ on $X$ (i.e. $fg= 0 \in A(X)$), and we can write $f(g\cdot 1 - f^n\cdot 0) = 0$.
Then $g/f^n\sim 0/1 \in A(X)_f$, which forces $g/f^n = 0\in A(X)_f$.
:::

## Presheaves and Sheaves

Idea: spaces on functions on topological spaces.

:::{.definition title="Presheaf"}
A *presheaf* (of rings) $\mathcal{F}$ on a topological space is

1. For every open set $U\subset  X$ a ring $\mathcal{F}(U)$.

2. For any inclusion $U\subset V$ a restriction map $\res_{VU}: \mathcal{F}(V) \to \mathcal{F}(U)$ satisfying

  a. $\mathcal{F}(\emptyset) = 0$.
  b. $\res_{UU} = \id_{\mathcal{F}(U)}$.
  c. $\res_{VW}  \circ \res_{UV} = \res_{UW}$.
:::


:::{.example}
The smooth functions on $\RR$ with the standard topology, $\mathcal{F} = C^\infty$ where $C^\infty(U)$ is the set of smooth functions $U\to \RR$.
It suffices to check the restriction condition, but the restriction of a smooth function is smooth: if $f$ is smooth on $U$, it is smooth at every point in $U$, i.e. all derivatives exist at all points of $U$. 
So if $V\subset U$, all derivatives of $f$ will exist at points $x \in V$, so $f$ will be smooth on $V$.

Note that this also works with continuous functions.
:::

:::{.definition title="Sheaf"}
A *sheaf* is a presheaf satisfying an additional gluing property: given $\phi_i \in \mathcal{F}(U_i)$ such that $\restrictionof{\phi_i}{U_i\intersect U_j} = \restrictionof{\phi_j}{U_i \intersect U_j}$, then there exists a unique $\phi\in \mathcal{F}(\union_i U_i)$ such that $\restrictionof{\phi}{U_i} = \phi_i$.
:::